Given:
The function \(\underline{r(t)=\ <\ 2\ \cos\ t,\ 2\ \sin\ t,\ t\ >}\)
Proofs:
The curve in terms of arc length is,
\(r(s)=2\ \cos\ \left(\frac{s}{\sqrt{5}}\right)i\ +\ 2\ \sin\ \left(\frac{s}{\sqrt{5}}\right)j\ +\ \frac{s}{\sqrt{5}}k\).
On differenting the vector-value function r(s), we get
\(r'(s)=\ -\frac{2}{\sqrt{5}}\ \sin \left(\frac{s}{\sqrt{5}}\right)i\ +\ \frac{2}{\sqrt{5}}\ \cos \left(\frac{s}{\sqrt{5}}\right)j\ +\ \frac{1}{\sqrt{5}}k\)
From this calcute \(\|r'(s)\|\) as
\(\|r'(s)\|=\ \sqrt{\left(-\frac{2}{\sqrt{5}}\ \sin\left(\frac{s}{\sqrt{5}}\right)\right)^{2}\ +\ \left(\frac{2}{\sqrt{5}}\ \cos\left(\frac{s}{\sqrt{5}}\right)\right)^{2}\ +\ \left(\frac{1}{\sqrt{5}}\right)^{2}}\)

\(=\ \sqrt{\frac{4}{5}\ +\ \frac{1}{5}}\)

\(=\ \sqrt{\frac{5}{5}}\)

\(= 1\) Hence, it is proved that \(\|r'(s)\|=1\)

\(=\ \sqrt{\frac{4}{5}\ +\ \frac{1}{5}}\)

\(=\ \sqrt{\frac{5}{5}}\)

\(= 1\) Hence, it is proved that \(\|r'(s)\|=1\)